Which of the following is a heterogeneous mixture?
A) vinegar and water
B) milk
C) Oil and vinegar
D) Air

Answer:

Hydrogen fuel cells create less pollution.

Burning fossil fuels relies on outdated devices and technology.

Hydrogen fuel cells have a higher energy efficiency.

Explanation:

Hello,

The first option - Hydrogen fuel cells create less pollution - is selected due to the fact that fuel cell consists of an electrolyte sandwiched  between two electrodes, an anode and a cathode, so electrical energy is used instead of the burning of a fossil fuel which gives off pollutant gases such as carbon, nitrogen and sulfur oxides.

The second option - Burning fossil fuels relies on outdated devices and technology - is selected due to even when new technologies are being developed to enhance the extraction of energy from the combustion of fossil fuels, they are based on the same idea, combustion which contributes to pollution.

The third option - Hydrogen fuel cells have a higher energy efficiency - is selected because a typical combustion process converts the 35% of the total fuel into usable energy meanwhile the hydrogen fuel cells reach the up to the 60% or more.

Best regards.

No, I believe this is not an example of a chemical reaction. What we actually see here is a physical change of the solution. Since we are adding more water to an aqueous solution which is also made up mostly of water, what we are simply basically doing is dilution. Since the solution is being diluted, so definitely the color turned lighter.

6.11×10²² oxygen atoms are present in 4.24 grams of ZnCO₃. The molar mass of ZnCO₃ is 125.34 g/mol.

Given information:

The molecular weight of ZnCO₃ = 125.34 g/mol

Number  of oxygen atoms = 6.11×10²²

1 mole of ZnCO₃ → 3 moles of oxygen atoms

1 mole of ZnCO₃ → atoms

125.34 g/mol ZnCO₃ → 3 × 6.023 × 10²³ atoms

Mass is a measure of the amount of matter in an object. It is a fundamental property of matter, and it does not change with the object's location or motion. The SI unit of mass is the kilogram (kg).

Now, the mass of ZnCO₃ is:

[tex]\rm Mass = \frac{6.11 \times 10^{22} \times 125.4}{3 \times 6.023 \times 10^{23}}\\\rm Mass = \frac{6.11 \times 10^{22} \times 125.4\times 10^{-23}}{3 \times 6.023 }\\Mass = \frac{766.194}{18.069}\times10^{-1}\\Mass = \frac{76.62}{18.069}\\[/tex]

Mass = 4.24 grams

Therefore, the mass of ZnCO₃ is 4.24 grams which contains 6.11×10²² oxygen atoms.

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To find the mass of ZnCO3 that contains 6.11×10^22 O atoms, we can calculate the molar mass of ZnCO3, use it to convert the number of oxygen atoms into moles, and then calculate the mass of ZnCO3. The mass of ZnCO3 is 12.73 g.

To find the mass of ZnCO3 that contains 6.11×10^22 O atoms, we need to calculate the molar mass of ZnCO3 and then use it to convert the number of oxygen atoms into moles. Finally, we can use the molar mass of ZnCO3 to calculate the mass of the compound.

The molar mass of ZnCO3 is calculated by summing the atomic masses of its constituent elements: Zn (65.38 g/mol), C (12.01 g/mol), and 3 O atoms (16.00 g/mol).

Molar mass of ZnCO3 = 65.38 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = 125.38 g/mol

Now, let's calculate the number of moles of O atoms:

Number of moles of O atoms = (6.11×10^22) / (6.022×10^23)

= 0.1015 mol

Finally, we can calculate the mass of ZnCO3:

Mass of ZnCO3 = (0.1015 mol) * (125.38 g/mol)

= 12.73 g

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The statement "each carbon-oxygen bond is somewhere between a single and double bond" and "the actual structure of formate is an average of the two resonance forms" are correct. The resonance structures represent an average distribution of electrons across all valid structures, not constant flipping between these structures.

In regards to the resonance structures of formate, the statements "each carbon–oxygen bond is somewhere between a single and double bond" and "the actual structure of formate is an average of the two resonance forms" are both true.

Resonance describes the situation where more than one valid Lewis structure can be drawn for a particular molecule. The resonance structure is not a rapid equilibrium between the structures but rather an average of the different possible structures, called resonance forms.

In the case of formate, which has two resonance forms, the molecule doesn't constantly flip between these two structures. Instead, the electrons are distributed in a way that is an average of these two resonance forms. This is why each carbon-oxygen bond in formate is described as being somewhere between a single and a double bond, as the characteristics of the bond are shared across both resonance forms.

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The mass of a sulfur-32 atom is approximately 8.01 x 10^-23g. There are approximately 5.97 x 10^23 neutrons in one gram.

The mass of a sulfur-32 atom, which is a type of nuclear atom, can be determined by summing the masses of its protons, neutrons, and electrons. Since a sulfur-32 atom has 16 protons, 16 neutrons, and 16 electrons, and given that the mass of a proton or neutron is approximately 1.67 x 10^-24 g, the total mass of the atom would be 48 x 1.67 x 10^-24g, which equals approximately 8.01 x 10^-23g.

Furthermore, to determine the number of neutrons that would equal a mass of one gram, we divide one gram by the mass of a single neutron. This results in approximately 5.97 x 10^23 neutrons.

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To identify the amount of chloride ions present in 0.1 mol of MgCl2, first know the amount of mols of Chloride. The answer is 0.2 mol since you just need to multiply 0.1 to 2 because there are 2 chloride in the compound. Then, multiply the mols to 6.02x10^23. The answer would be 1.204x10^23 chloride ions. 

Answer : The number of chloride ions present in 0.100 mole of [tex]MgCl_2[/tex] are, [tex]1.2044\times 10^{23}[/tex]

Explanation : Given,

Moles of [tex]MgCl_2[/tex] = 0.100 mole

As we know that, 1 mole contains [tex]6.022\times 10^{23}[/tex] number of ions.

In [tex]MgCl_2[/tex], there are one magnesium ion and two chloride ions.

As, 1 mole [tex]MgCl_2[/tex] contains [tex]2\times (6.022\times 10^{23})[/tex] number of chloride ions.

So, 0.100 mole [tex]MgCl_2[/tex] contains [tex]0.100\times 2\times (6.022\times 10^{23})=1.2044\times 10^{23}[/tex] number of chloride ions.

Therefore, the number of chloride ions present in 0.100 mole of [tex]MgCl_2[/tex] are, [tex]1.2044\times 10^{23}[/tex]

The ionic compound formed from Fe3+ and Cl- is FeCl3, called iron(III) chloride. For dissociation in water, FeCl2 separates into Fe2+ and Cl-. The complete ionic equation for the reaction between FeCl2 and AgNO3 includes all ions present with AgCl precipitating out, and the net ionic equation shows just the ions involved in the formation of the precipitate.

Naming the Ionic Compound Formed from Fe3+ and Each Anion

To name the ionic compound formed from Fe3+ and an anion, you follow a set of rules. Firstly, you write the name of the cation including the charge in roman numerals enclosed in small brackets, followed by the name of the anion. For Fe3+, we use the name Iron(III), and for a chloride ion (Cl-), the name is chloride. The compound's formula is FeCl3, which is called iron(III) chloride. Remember that the subscript in the chemical formula indicates the number of anions needed to balance the charge of the cation.

Compound Dissociation in Water

To write a balanced equation for how an ionic compound like FeCl2 dissociates in water, you would represent it as follows:

FeCl2(s) -> Fe2+(aq) + 2 Cl-(aq)

This shows the separation of the ionic compound into its individual ions when dissolved in water.

Complete Ionic Equation for FeCl2 and AgNO3

The complete ionic equation for the reaction between FeCl2(aq) and AgNO3(aq), consulting solubility rules, would look like this:

Fe2+(aq) + 2 Cl-(aq) + 2 Ag+(aq) + 2 NO3-(aq) -> 2 AgCl(s) + Fe2+(aq) + 2 NO3-(aq)

Net Ionic Equation for FeCl2 and AgNO3

The net ionic equation, after removing the spectator ions, for the same reaction would be:

2 Cl-(aq) + 2 Ag+(aq) -> 2 AgCl(s)

The number of Li atoms in 5.1 moles of Li is 3.07 x 10²⁴ atoms.

Atoms are defined as the smallest piece of matter that can be separated without sending electrically charged particles flying.

It can also be defined as the smallest piece that carries an element's characteristics. Subatomic particles, which make up an atom, are uncreatable.

There are various types of atoms.

In a chemical reaction, atoms cannot be formed or destroyed since they are indivisible units. The mass and chemical characteristics of each atom of a specific element are the same. Different elements' atoms have varying weight and chemical characteristics. Compounds are created when atoms combine in ratios of small whole numbers.

Moles of Li = 5.1 x 6.022 x 10²³

                   = 3.07 x 10²⁴ atoms.

Thus, the number of Li atoms in 5.1 moles of Li is 3.07 x 10²⁴ atoms.

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To calculate the molality of oxygen in water under a partial pressure of 0.20 atm, apply Henry's law with the given constant (773 atm mol⁻¹kg H₂O), resulting in a molality of 154.6 mol kg⁻¹.

The question asks to calculate the molality of oxygen in water at a specific temperature (25 °C) and partial pressure, using the given Henry's law constant. To find the molality, we need to apply Henry's law, which states that the concentration of a dissolved gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.

Hence, the formula for Henry's law is:

C = kP

where:

C is the molar concentration of the dissolved gas (molality in this case),

k is the Henry's law constant,

P is the partial pressure of the gas above the liquid.

Using the given Henry's law constant for oxygen in water (773 atm mol⁻¹kg H₂O) and the partial pressure of oxygen (0.20 atm), we can solve for C as follows:

C = kP

C = 773 atm mol⁻¹kg H₂O × 0.20 atm

C = 154.6 mol kg⁻¹

So, the molality of oxygen in water under a partial pressure of 0.20 atm is 154.6 mol kg⁻¹.

Answer : The number of moles of phosphorous are 354 moles.

Explanation :

The formula of given compound is, [tex]P_4O_{10}[/tex]

In [tex]P_4O_{10}[/tex] compound, there 4 moles of phosphorus and 10 moles of oxygen.

As we are given that the moles of [tex]P_4O_{10}[/tex] is 88.5 mole. Now we have to determine the number of moles of phosphorous (P).

As, 1 mole of [tex]P_4O_{10}[/tex] has 4 moles of phosphorous

So, 88.5 mole of [tex]P_4O_{10}[/tex] has [tex]4\times 88.5=354moles[/tex] of phosphorous

Therefore, the number of moles of phosphorous are 354 moles.

88.5 mol of [tex]P_4O_{10}[/tex] contains 354 moles of P. A mole is a unit of measurement used in chemistry to represent how much of a substance is present.

In chemistry, a mole is a unit of measurement that is used to express how much a material is present. It is described as the quantity of a substance that has exactly the same number of atoms, molecules, or ions as there are in exactly 12 grammes of pure carbon-12. The mole idea is essential to chemistry because it enables researchers to connect a substance's mass to its particle count. The idea of molar mass, or the mass of one mole of a substance, is used to describe this relationship. The unit of molecular mass is grammes per mole (g/mol).

88.5 mol [tex]P_4O_{10}[/tex] × (4 mol P / 1 mol P4O10) = 354 mol P

88.5 mol of [tex]P_4O_{10}[/tex] contains 354 moles of P

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The final temperature at thermal equilibrium can be calculated using the concept of conservation of energy and the specific heat capacities of gold and water.

To find the final temperature of the gold and water system when they reach thermal equilibrium, we need to apply the concept of conservation of energy. This concept suggests that in an isolated system, the heat lost by the hot object (the gold) will be equal to the heat gained by the cold object (the water). Since the system is at equilibrium, the heat lost is equal to the heat gained, hence the formula: Cgold × mgold × (Tinitial, gold - Tfinal) = -Cwater × mwater × (Tfinal - Tinitial, water), where Cgold and Cwater are the specific heat capacities of gold and water, T is the temperature and m is the mass.

We also need to know the specific heat capacities of gold and water. The specific heat capacity of gold is 0.129 J/g °C and for water, it's 4.18 J/g °C. Substituting those values along with the original temperatures and masses, we can solve for the final temperature, Tfinal.

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Actually, a methylene group is simply any compound which contains a C=C double bond group and the rest are single bonded carbon groups. Some example of the isomers of C6H12 which contains methylene group is:

1-hexene

2,3-dimethyl-2-butene

2,3-dimethyl-1-butene

2-methyl-2-pentene

trans-2-hexene

4-methyl-1-pentene

cis-2-hexene

trans-3-hexene

2-ethyl-1-butene

2-methyl-1-pentene

3,3-dimethyl-1-butene

4-methyl-cis-2-pentene

cis-3-methyl-2-pentene

trans-3-methyl-2-pentene

The empirical formula of the hydrocarbon is [tex]\boxed{{{\text{C}}_7}{{\text{H}}_5}}[/tex].

Further explanation:

Empirical formula:

It is atom’s simplest positive integer ratio in the compound. It may or may not be same as that of molecular formula. For example, empirical formula of sulfur dioxide is SO.

Combustion reactions:

These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.  

Example of combustion reactions are as follows:

(a) [tex]{\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}[/tex]

(b) [tex]{{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+ 4{{\text{H}}_2}{\text{O}}[/tex]

[tex]{\text{C}}{{\text{O}}_2}[/tex] is formed as a product during combustion reactions.

Step 1: [tex]{\text{C}}{{\text{O}}_2}[/tex] is formed as a product during combustion reactions. Initially, we have to calculate the moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] . The formula to calculate moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is as follows:

[tex]{\text{Moles of C}}{{\text{O}}_2}=\dfrac{{{\text{Given mass of C}}{{\text{O}}_2}}}{{{\text{Molar mass of C}}{{\text{O}}_2}}}[/tex]          ...... (1)

The given mass of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 86.5 g.

The molar mass of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 44 g/mol.

Substitute these values in equation (1).

[tex]\begin{aligned}{\text{Moles of C}}{{\text{O}}_2}&=\left( {{\text{86}}{\text{.5 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{44 g}}}}} \right)\\&={\text{1}}{\text{.9659 mol}}\\ &\approx{\text{1}}{\text{.966 mol}}\\\end{aligned}[/tex]

Step 2: During combustion, one mole of carbon reacts to form one mole of  [tex]{\text{C}}{{\text{O}}_2}[/tex] .So the mass of C in the hydrocarbon is calculated as follows:

[tex]{\text{Mass of C}}=\left( {{\text{Moles of C}}{{\text{O}}_{\text{2}}}}\right)\left( {\dfrac{{{\text{Moles of C}}}}{{{\text{Moles of C}{{\text{O}}_{\text{2}}}}}}\right)\left( {{\text{Molar mass of C}}}\right)[/tex]      ...... (2)

The moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 1.966 mol.

The molar mass of C is 12 g/mol.

The mole of C is 1 mol.

The moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 1 mol.

Substitute these values in equation (2).

[tex]\begin{aligned}{\text{Mass of C}}&=\left( {{\text{1}}{\text{.966 mol}}} \right)\left( {\frac{{{\text{1 mol of C}}}}{{{\text{1 mol of C}}{{\text{O}}_{\text{2}}}}}} \right)\left( {\frac{{{\text{12 g}}}}{{{\text{1 mol}}}}} \right)\\&= {\text{23}}{\text{.592 g}}\\&\approx{\text{23}}{\text{.59 g}}\\\end{aligned}\\[/tex]

Step 3: Since the hydrocarbon consists of only carbon and hydrogen. The mass of hydrogen is calculated as follows:

[tex]{\text{Mass of H}}={\text{Mass of hydrocarbon}}-{\text{Mass of C}}[/tex]      ...... (3)

The mass of hydrocarbon is 25 g.

The mass of carbon is 23.59 g.

Substitute these values in equation (3).

[tex]\begin{aligned}{\text{Mass of H}}&={\text{25 g}}-{\text{23}}{\text{.59 g}}\\&= {\text{1}}{\text{.41 g}}\\\end{aligned}[/tex]

The formula to calculate moles of H is as follows:[tex]{\text{Moles of H}}=\dfrac{{{\text{Given mass of H}}}}{{{\text{Molar mass of H}}}}[/tex]      ...... (4)

The given mass of H is 1.41 g.

The molar mass of H is 1.01 g/mol.

Substitute these values in equation (4).

[tex]\begin{aligned}{\text{Moles of H}}&=\left({{\text{1}}{\text{.41 g}}}\right)\left( {\frac{{{\text{1 mol}}}}{{{\text{1}}{\text{.01 g}}}}}\right)\\&={\text{1}}{\text{.396 mol}}\\ \end{aligned}[/tex]

The moles of carbon and hydrogen present in hydrocarbon are to be written with their corresponding subscripts. So the preliminary formula becomes,

[tex]{\text{Preliminary formula of hydrocarbon}}={{\text{C}}_{1.966}}{{\text{H}}_{1.396}}[/tex]

Step 4: Each of the subscripts is divided by the smallest subscript to get the empirical formula. In this case, the smallest one is 1.39. So the empirical formula of hydrocarbon is written as follows:

[tex]\begin{aligned}{\text{Empirical formula of hydrocarbon}}&={{\text{C}}_{\dfrac{{1.966}}{{1.396}}}}{{\text{H}}_{\dfrac{{1.396}}{{1.396}}}}\\&= {{\text{C}}_{1.408}}{{\text{H}}_1}\\\end{aligned}[/tex]

Step 5: Multiply each subscript of the empirical formula by 5, we get the final empirical formula as follows:

[tex]\begin{aligned}{\text{Empirical formula of hydrocarbon}}&={{\text{C}}_{5\left( {1.408} \right)}}{{\text{H}}_{5\left( 1 \right)}}\\&={{\text{C}}_7}{{\text{H}}_5}\\\end{aligned}[/tex]

Therefore, the empirical formula of hydrocarbon is [tex]{{\mathbf{C}}_{\mathbf{7}}}{{\mathbf{H}}_{\mathbf{5}}}[/tex] .

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1. Calculate the moles of ions in the solution: https://brainly.com/question/5950133

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Stoichiometry of formulas and equations

Keywords: empirical formula, C, H, C7H5, moles of CO2, C, H, 5, preliminary formula, whole number.

the final temperature of the water is approximately [tex]\(4.009 C\)[/tex].

To find the final temperature of the water after adding the heat produced from burning 8.800 g of [tex]\(C_6H_6\)[/tex] (benzene), we'll use the concept of heat transfer and the specific heat capacity of water.

The heat released from the combustion of [tex]\(C_6H_6\)[/tex] will be transferred to the water, causing its temperature to increase. We'll use the equation:

Q = mcΔT

Where:

- Q is the heat transferred (in Joules)

- m is the mass of the water (in grams)

- c is the specific heat capacity of water (4.18 J/g°C)

- ΔT  is the change in temperature of the water (in °C)

First, we need to calculate the heat released from burning [tex]\(C_6H_6\)[/tex].

Given:

- Mass of [tex]\(C_6H_6\)[/tex] burned, [tex]\(m_{C_6H_6} = 8.800 \, g\)[/tex]

- Heat of combustion of [tex]\(C_6H_6\)[/tex], [tex]\(ΔH_{comb} = -3263 \, kJ/mol\)[/tex]

Using the molar mass of [tex]\(C_6H_6\) (\(M_{C_6H_6} = 78.11 \, g/mol\))[/tex], we can find the number of moles of [tex]\(C_6H_6\)[/tex] burned and then calculate the heat released.

Next, we'll use the equation for heat transfer to find the change in temperature of the water, and then add this change to the initial temperature of the water to get the final temperature.

Let's calculate step by step.

Step 1: Calculate the heat released from burning [tex]\(C_6H_6\)[/tex].

1. Find the number of moles of [tex]\(C_6H_6\)[/tex]:

[tex]\[n_{C_6H_6} = \frac{m_{C_6H_6}}{M_{C_6H_6}} = \frac{8.800 \, g}{78.11 \, g/mol} \approx 0.1128 \, mol\][/tex]

2. Calculate the heat released from burning [tex]\(C_6H_6\)[/tex] using its molar enthalpy of combustion:

[tex]\[Q_{comb} = n_{C_6H_6} \times ΔH_{comb} = 0.1128 \, mol \times (-3263 \, kJ/mol)\][/tex]

[tex]\[Q_{comb} = -368.112 \, kJ\][/tex]

Step 2: Calculate the change in temperature of the water.

1. Use the equation for heat transfer:

[tex]\[Q_{water} = mcΔT\][/tex]

Where [tex]\(Q_{water}\)[/tex] is the heat absorbed by water, \(m\) is the mass of water, c is the specific heat capacity of water, and \(ΔT\) is the change in temperature of water.

2. Rearrange the equation to solve for [tex]\(ΔT\)[/tex]:

[tex]\[ΔT = \frac{Q_{comb}}{mc}\][/tex]

Given:

- [tex]\(m_{water} = 5691 \, g\)[/tex]

- [tex]\(c_{water} = 4.18 \, J/g°C\)[/tex]

3. Substitute the values and calculate \(ΔT\):

[tex]\[ΔT = \frac{-368.112 \times 10^3 \, J}{(5691 \, g) \times (4.18 \, J/g°C)}\][/tex]

[tex]\[ΔT \approx -16.991°C\][/tex]

Step 3: Find the final temperature of the water.

Given:

- Initial temperature of water, [tex]\(T_{initial} = 21°C\)[/tex]

The final temperature [tex](\(T_{final}\))[/tex] of the water can be found by adding the change in temperature [tex](\(ΔT\))[/tex] to the initial temperature [tex](\(T_{initial}\))[/tex]:

[tex]\[T_{final} = T_{initial} + ΔT\][/tex]

[tex]\[T_{final} = 21°C - 16.991°C\][/tex]

[tex]\[T_{final} \approx 4.009°C\][/tex]

Therefore, the final temperature of the water is approximately [tex]\(4.009 C\)[/tex].

beryllium and magnesium is your answer

The two elements which have the most similar chemical properties are: Beryllium and Magnesium.

Discussion:

The two elements are present in the Alkali earth metals group of the periodic table.

They are both characterized by the possession of 2 Valence electrons on their outermost shell and as such are able to undergo similar chemical reactions.

Ultimately, beryllium and magnesium have the most similar chemical properties.

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Answer:The pH of the water is 7.

Explanation;

The pH of the solution is defined as negative logarithm of [tex]H^+[/tex] or hydronium ions ions in the solution.

[tex]pH=-\log[H^+][/tex]

So, [tex]H^+[/tex] concentration of water = [tex]1.0\times 10^{-7} m[/tex]

[tex]pH=-\log[1.0\times 10^{-7}]=7[/tex]

The pH of the water is 7.

Answer : The density of an object is 1.2 g/mL

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used :

[tex]Density=\frac{Mass}{Volume}[/tex]

Given :

Mass of object = 0.03 kg = 30 g

conversion used : (1 kg = 1000 g)

Volume occupied by object = 25 mL

Now put all the given values in the above formula, we get :

[tex]Density=\frac{30g}{25mL}=1.2g/mL[/tex]

Therefore, the density of an object is 1.2 g/mL

The average atomic mass of lead, will be approximately 207.2 g/mol.

To calculate the approximate atomic mass of lead using the isotopic masses and their relative abundances. The atomic mass is a weighted average based on the abundances of each isotope.

For lead (Pb), the calculation using given isotopes would look like this:

Thus, the approximate atomic mass of Pb is 207.2 g/mol.

The complete question is:

The four isotopes of lead are shown below, each with its percent by mass abundance and the composition of its nucleus. using these data, calculate the approximate atomic mass of lead.  82p 122n 1.37% 82p 124n 26.26% 125n 20.82 82p 126n 51.55% 24 Mass # 81 37 6.

To precipitate all the bromide ions present in the final solution of KBr, 12.061g of silver nitrate is needed. The calculation is based on molarity, volume and reaction stoichiometry.

The first step to finding the answer is to calculate the amount of KBr in each solution. The amount of solute in a solution is given by the formula: Volume (L) × Molarity (M). So for the 35.0 mL sample of 1.00 M KBr, that would be 0.035 L × 1.00 mol/L = 0.035 moles of KBr. For the 60.0 mL sample of 0.600 M KBr, the calculation is 0.060 L × 0.600 mol/L = 0.036 moles of KBr. Thus, a total of 0.035 + 0.036 = 0.071 moles KBr is present in the final solution.

The reaction between KBr and silver nitrate (AgNO3) is a one-to-one reaction: KBr + AgNO3 → AgBr + KNO3. Therefore, 0.071 moles of AgNO3 are required to precipitate all the bromide ions. The molar mass of AgNO3 is approximately 169.87 g/mol, therefore the total mass of AgNO3 needed is 0.071 moles × 169.87 g/mol = 12.061 g.

Hence, 12.061 g of silver nitrate are required to precipitate out silver bromide in the final solution.

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The amount of silver nitrate required is approximately 10.53 grams. This is calculated by first determining the total moles of KBr in the solution and then using the 1:1 stoichiometry of the reaction to find the equivalent moles (and thus mass) of AgNO3 required.

This is a stoichiometry problem that involves determining the amount of silver nitrate needed to precipitate out silver bromide in a solution. First, we need to determine the number of moles of KBr in the final solution. The original solutions have volumes of 35.0 ml and 60.0 ml and molarities of 1.00 M and 0.600 M respectively, so the number of moles of KBr is (35.0 ml x 1.00 mol/L) + (60.0 ml x 0.600 mol/L) = 0.062 mol. After evacuating some water, the volume decreases to 50.0 ml.

Now, since silver nitrate and potassium bromide react in a 1:1 ratio to form silver bromide, the moles of silver nitrate needed will be equal to the moles of KBr. Thus, we need 0.062 moles of AgNO3. The molecular weight of AgNO3 is 169.87 g/mol, so the mass of AgNO3 needed is (0.062 mol) x (169.87 g/mol) = 10.53 g. So, you will need about 10.53 grams of silver nitrate to precipitate out silver bromide in this solution.

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